Chapter 16.4, Problem 21E

To determine

(a)

To show:

${\int }_C^{\square }x dy-y dx=x_1{y}_2-x_2{y}_1$

If $C$ is line segment connecting point $\left(x_1, y_1\right)$ to the point $(x_2,y_2)$

Solution:

${\int }_C^{\square }x dy-y dx=x_1{y}_2-x_2{y}_1$

Explanation:

1) Concept:

The area of region $D$ is

$A= {\oint }_{C }^{\square }x dy= -{\oint }_{C }^{\square }y dx=\frac{1}{2}{\oint }_{C }^{\square } x dy-y dx$

2) Given:

The line segment $C$ connecting point $\left(x_1, y_1\right)$ to the point $(x_2,y_2)$

3) Calculation:

The parametric equation for line segmentconnecting point $\left(x_1, y_1\right)$ to the point $(x_2,y_2)$ is

$x=x_1+t\left(x_2-x_1\right), y=y_1+t\left(y_2-y_1\right) , 0\le t\le 1$

$dx=\left(x_2-x_1\right) dt$ and $dy=\left(y_2-y_1\right) dt$

${\oint }_{C }^{\square } x dy-y dx= {\int }_0^1[x_1+t\left(x_2-x_1\right)\left] \right(y_2-y_1) dt-[y_1+t\left(y_2-y_1\right)\left] \right(x_2-x_1) dt$

$= {\int }_0^1\left(x_1\left(y_2-y_1\right)-y_1\left(x_2-x_1\right)+t \left[\left(y_2-y_1\right)\left(x_2-x_1\right)-\left(x_2-x_1\right)\left(y_2-y_1\right)\right]\right) dt$

$= {\int }_0^1 (\left(x_1{y}_2-x_1{y}_1-{ x}_2{y}_1+x_1{y}_1\right)+t \left(0\right) ) dt$

$= {\int }_0^1\left(x_1{y}_2-x_2{y}_1\right)dt$

$=x_1{y}_2-x_2{y}_1$

Hence proved

Conclusion:

${\int }_C^{\square }x dy-y dx=x_1{y}_2-x_2{y}_1$

(b)

To show:

That the area of the polygon is

$A=\frac{1}{2}\left[\left(x_1{y}_2-x_2{y}_1\right)+\left(x_2{y}_3-x_3{y}_2\right)+\dots +\left(x_{n-1}{y}_n-x_n{y}_{n-1}\right)+\left(x_n{y}_1-x_1{y}_n\right)\right]$

Solution:

$A=\frac{1}{2}\left[\left(x_1{y}_2-x_2{y}_1\right)+\left(x_2{y}_3-x_3{y}_2\right)+\dots +\left(x_{n-1}{y}_n-x_n{y}_{n-1}\right)+\left(x_n{y}_1-x_1{y}_n\right)\right]$

Explanation:

1) Concept:

By Green’s Theorem the area of region $D$ is

$A= {\oint }_{C }^{\square }x dy= -{\oint }_{C }^{\square }y dx=\frac{1}{2}{\oint }_{C }^{\square } x dy-y dx$

2) Given:

The vertices of polygon in a counterclockwise order are $\left(x_1, y_1\right), \left(x_2,y_2\right),\dots , \left(x_n,y_n\right)$

3) Calculation:

Apply Green’s Theorem to the path $C= C_1 \cup C_2\cup \dots \cup C_n$

where $C_i$ is line segment that joins $\left(x_i, y_i\right) to \left(x_{i+1},y_{i+1}\right)$ for $i=1, 2, \dots , n-1$ and

$C_n$ is line segment that joins $\left(x_n,y_n\right)$ to $\left(x_1, y_1\right)$

Using the concept,

Area of polygon,

$A\left(D\right)= {\iint }_D^{\square }dA=\frac{1}{2}{\int }_C^{\square }x dy-y dx$

$=\frac{1}{2}{\int }_{C_1}^{\square }x dy-y dx+ {\int }_{C_2}^{\square }x dy-y dx+\dots + {\int }_{C_{n-1}}^{\square }x dy-y dx+{\int }_{C_n}^{\square }x dy-y dx$

Using the formula in part (a)

$A=\frac{1}{2}\left[\left(x_1{y}_2-x_2{y}_1\right)+\left(x_2{y}_3-x_3{y}_2\right)+\dots +\left(x_{n-1}{y}_n-x_n{y}_{n-1}\right)+\left(x_n{y}_1-x_1{y}_n\right)\right]$

Conclusion:

$A=\frac{1}{2}\left[\left(x_1{y}_2-x_2{y}_1\right)+\left(x_2{y}_3-x_3{y}_2\right)+\dots +\left(x_{n-1}{y}_n-x_n{y}_{n-1}\right)+\left(x_n{y}_1-x_1{y}_n\right)\right]$

(c)

To find:

The area of pentagon with the given vertices

Solution:

$A =\frac{9}{2}$

Explanation:

1) Concept:

Formula from part (b)

$A=\frac{1}{2}\left[\left(x_1{y}_2-x_2{y}_1\right)+\left(x_2{y}_3-x_3{y}_2\right)+\dots +\left(x_{n-1}{y}_n-x_n{y}_{n-1}\right)+\left(x_n{y}_1-x_1{y}_n\right)\right]$

2) Given:

The pentagon with vertices $(0, 0), (2, 1), (1, 3), (0, 2), (-1, 1)$

3) Calculation:

Using the formula, Area is

$A=\frac{1}{2}\left[\left(0·1-2.0\right)+\left(2·3-1·1\right)+\left(1·2+0·3\right)+\left(0·1-\left(-1\right)·2\right)+\left(-1·0-0·1l\right)\right]$

$=\frac{1}{2}\left(0+5+2+2+0\right)=\frac{9}{2}$

Thus, area is $\frac{9}{2}$

Conclusion:

$A =\frac{9}{2}$